1352. Product of the Last K Numbers

 

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

 

Example:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

 

Restrictions:

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

 

这道题是之前那道题Reverse String的拓展，同样是翻转字符串，但是这里是每隔k隔字符，翻转k个字符，最后如果不够k个了的话，剩几个就翻转几个。比较直接的方法就是先用n／k算出来原字符串s能分成几个长度为k的字符串，然后开始遍历这些字符串，遇到2的倍数就翻转，翻转的时候注意考虑下是否已经到s末尾了，参见代码如下：

 

解法一：

class Solution {
public:
    string reverseStr(string s, int k) {
        int n = s.size(), cnt = n / k;
        for (int i = 0; i <= cnt; ++i) {
            if (i % 2 == 0) {
                if (i * k + k < n) {
                    reverse(s.begin() + i * k, s.begin() + i * k + k);
                } else {
                    reverse(s.begin() + i * k, s.end());
                }
            }
        }
        return s;
    }
};

 

在论坛里又发现了写法更为简洁的方法，就是每2k个字符来遍历原字符串s，然后进行翻转，翻转的结尾位置是取i+k和末尾位置之间的较小值，感觉很叼，参见代码如下：

 

解法二：

class Solution {
public:
    string reverseStr(string s, int k) {
        for (int i = 0; i < s.size(); i += 2 * k) {
            reverse(s.begin() + i, min(s.begin() + i + k, s.end()));
        }
        return s;
    }
};

 

类似题目：

Reverse String

 

参考资料：

https://discuss.leetcode.com/topic/82652/one-line-c/2

https://discuss.leetcode.com/topic/82626/java-concise-solution

 

LeetCode All in One 题目讲解汇总(持续更新中…)

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